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5.5: Definite Integral Intro Exercises

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5.2: The Definite Integral

In the following exercises, express the limits as integrals.

60) limnni=1(xi)Δx over [1,3]

61) limnni=1(5(xi)23(xi)3)Δx over [0,2]

Answer:
20(5x23x3)dx

62) limnni=1sin2(2πxi)Δx over [0,1]

63) limnni=1cos2(2πxi)Δx over [0,1]

Answer:
10cos2(2πx)dx

In the following exercises, given Ln or Rn as indicated, express their limits as n as definite integrals, identifying the correct intervals.

64) Ln=1nni=1i1n

65) Rn=1nni=1in

Answer:
fave=4528

Solution: 10xdx

66) Ln=2nni=1(1+2i1n)

67) Rn=3nni=1(3+3in)

Solution: 63xdx

68) Ln=2πnni=12πi1ncos(2πi1n)

69 Rn=1nni=1(1+in)log((1+in)2)

Solution: 21xlog(x2)dx

In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the x-axis.

70)

CNX_Calc_Figure_05_02_201.jpeg

71)

CNX_Calc_Figure_05_02_202.jpeg

Solution: 1+22+33=14

72)

CNX_Calc_Figure_05_02_203.jpeg

73)

CNX_Calc_Figure_05_02_204.jpeg

Solution: 14+9=6

74)

CNX_Calc_Figure_05_02_205.jpeg

75)

CNX_Calc_Figure_05_02_206.jpeg

Solution: 12π+9=102π

In the following exercises, evaluate the integral using area formulas.

76) 30(3x)dx

77) 32(3x)dx

Answer:
The integral is the area of the triangle, 12

78) 33(3|x|)dx

79) 60(3|x3|)dx

Solution: The integral is the area of the triangle, 9.

80) 224x2dx

81) 514(x3)2dx

Soluti0n: The integral is the area 12πr2=2π.

82) 12036(x6)2dx

83) 32(3|x|)dx

Solution: The integral is the area of the “big” triangle less the “missing” triangle, 912.

In the following exercises, use averages of values at the left (L) and right (R) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals.

84) (0,0),(2,1),(4,3),(5,0),(6,0),(8,3) over [0,8]

85) (0,2),(1,0),(3,5),(5,5),(6,2),(8,0) over [0,8]

Solution: L=2+0+10+5+4=21,R=0+10+10+2+0=22,L+R2=21.5

86) (4,4),(2,0),(0,2),(3,3),(4,3) over [4,4]

87) (4,0),(2,2),(0,0),(1,2),(3,2),(4,0) over [4,4]

Solution: L=0+4+0+4+2=10,R=4+0+2+4+0=10,L+R2=10

Suppose that 40f(x)dx=5 and 20f(x)dx=3, and 40g(x)dx=1 and 20g(x)dx=2. In the following exercises, compute the integrals.

88) 40(f(x)+g(x))dx

89) 42(f(x)+g(x))dx

Answer:
42f(x)dx+42g(x)dx=83=5

90) 20(f(x)g(x))dx

91) 42(f(x)g(x))dx

Answer:
42f(x)dx42g(x)dx=8+3=11

92) 20(3f(x)4g(x))dx

93) 42(4f(x)3g(x))dx

Answer:
442f(x)dx342g(x)dx=32+9=41

In the following exercises, use the identity AAf(x)dx=0Af(x)dx+A0f(x)dx to compute the integrals.

94) ππsint1+t2dt (Hint: sin(t)=sin(t))

95) ππt1+costdt

Solution: The integrand is odd; the integral is zero.

96) 31(2x)dx (Hint: Look at the graph of f.)

97) 42(x3)3dx (Hint: Look at the graph of f.)

Solution: The integrand is antisymmetric with respect to x=3. The integral is zero.

In the following exercises, given that 10xdx=12,10x2dx=13, and 10x3dx=14, compute the integrals.

98) 10(1+x+x2+x3)dx

99) 10(1x+x2x3)dx

Answer:
112+1314=712

100) 10(1x)2dx

101) 10(12x)3dx

Solution: 10(12x+4x28x3)dx=11+432=23

102) 10(6x43x2)dx

103) 10(75x3)dx

Solution: 754=234

In the following exercises, use the comparison theorem.

104) Show that 30(x26x+9)dx0.

105) Show that 32(x3)(x+2)dx0.

Solution: The integrand is negative over [2,3].

106) Show that 101+x3dx101+x2dx.

107) Show that 211+xdx211+x2dx.

Solution: xx2 over [1,2], so 1+x1+x2 over [1,2].

108) Show that π/20sintdtπ4) (Hint: sint2tπ over [0,π2])

109) Show that π/4π/4costdtπ2/4.

Solution: cos(t)22. Multiply by the length of the interval to get the inequality.

In the following exercises, approximate the average value using Riemann sums L100 and R100. How does your answer compare with the exact given answer?

116) [T] y=ln(x) over the interval [1,4]; the exact solution is ln(256)31.

117) [T] y=ex/2 over the interval [0,1]; the exact solution is 2(e1).

5olution: L100=1.294,R100=1.301; the exact average is between these values.

118) [T] y=tanx over the interval [0,π4]; the exact solution is 2ln(2)π.

119) [T] y=x+14x2 over the interval [1,1]; the exact solution is π6.

Solution: L100×(12)=0.5178,R100×(12)=0.5294

In the following exercises, compute the average value using the left Riemann sums LN for N=1,10,100. How does the accuracy compare with the given exact value?

120) [T] y=x24 over the interval [0,2]; the exact solution is 83.

121) [T] y=xex2 over the interval [0,2]; the exact solution is 14(e41).

Solution: L1=0,L10×(12)=8.743493,L100×(12)=12.861728. The exact answer 26.799, so L100 is not accurate.

122) [T] y=(12)x over the interval [0,4]; the exact solution is 1564ln(2).

123) [T] y=xsin(x2) over the interval [π,0]; the exact solution is cos(π2)12π.

Solution: L1×(1π)=1.352,L10×(1π)=0.1837,L100×(1π)=0.2956. The exact answer 0.303, so L100 is not accurate to first decimal.

124) Suppose that A=2π0sin2tdt and B=2π0cos2tdt. Show that A+B=2π and A=B.

125) Suppose that A=π/4π/4sec2tdt=π and B=π/4π/4tan2tdt. Show that AB=π2.

Solution: Use tan2θ+1=sec2θ. Then, BA=π/4π/41dx=π2.

126) Show that the average value of sin2t over [0,2π] is equal to 1/2 Without further calculation, determine whether the average value of sin2t over [0,π] is also equal to 1/2.

127) Show that the average value of cos2t over [0,2π] is equal to 1/2. Without further calculation, determine whether the average value of cos2(t) over [0,π] is also equal to 1/2.

Solution: 2π0cos2tdt=π, so divide by the length 2π of the interval. cos2t has period π, so yes, it is true.

128) Explain why the graphs of a quadratic function (parabola) p(x) and a linear function ℓ(x) can intersect in at most two points. Suppose that p(a)=(a) and p(b)=(b), and that bap(t)dt>ba(t)dt. Explain why dcp(t)>dc(t)dt whenever ac<db.

129) Suppose that parabola p(x)=ax2+bx+c opens downward (a<0) and has a vertex of y=b2a>0. For which interval [A,B] is BA(ax2+bx+c)dx as large as possible?

Solution: The integral is maximized when one uses the largest interval on which p is nonnegative. Thus, A=bb24ac2a and B=b+b24ac2a.

130) Suppose [a,b] can be subdivided into subintervals a=a0<a1<a2<<aN=b such that either f0 over [ai1,ai] or f0 over [ai1,ai]. Set Ai=aiai1f(t)dt.

a. Explain why baf(t)dt=A1+A2++AN.

b. Then, explain why baf(t)dt∣∣∣≤ba|f(t)|dt.

131) Suppose f and g are continuous functions such that dcf(t)dtdcg(t)dt for every subinterval [c,d] of [a,b]. Explain why f(x)g(x) for all values of x.

Solution: If f(t0)>g(t0) for some t0[a,b], then since fg is continuous, there is an interval containing t0 such that f(t)>g(t) over the interval [c,d], and then ddf(t)dt>dcg(t)dover this interval.

132) Suppose the average value of f over [a,b] is 1 and the average value of f over [b,c] is 1 where a<c<b. Show that the average value of f over [a,c] is also 1.

133) Suppose that [a,b] can be partitioned. taking a=a0<a1<<aN=b such that the average value of f over each subinterval [ai1,ai]=1 is equal to 1 for each i=1,,N. Explain why the average value of f over [a,b] is also equal to 1.

Solution: The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, baf(t)dt=a1a0f(t)dt+a2a1f(t)dt++aNaN+1f(t)dt=a1a01dt+a2a11dt++aNaN+11dt

=(a1a0)+(a2a1)++(aNaN1)=aNa0=ba.

Dividing through by ba gives the desired identity.

134) Suppose that for each i such that 1iN one has ii1f(t)dt=i. Show that N0f(t)dt=N(N+1)2.

135) Suppose that for each i such that 1iN one has ii1f(t)dt=i2. Show that N0f(t)dt=N(N+1)(2N+1)6.

Solution: N0f(t)dt=Ni=1ii1f(t)dt=Ni=1i2=N(N+1)(2N+1)6

136) [T] Compute the left and right Riemann sums L10 and R10 and their average L10+R102 for f(t)=t2over [0,1]. Given that 10t2dt=1/3, to how many decimal places is L10+R102 accurate?

137) [T] Compute the left and right Riemann sums, L10 and R10, and their average L10+R102 for f(t)=(4t2) over [1,2]. Given that 21(4t2)dt=1.66, to how many decimal places is L10+R102 accurate?

Solution: L10=1.815,R10=1.515,L10+R102=1.665, so the estimate is accurate to two decimal places.

138) If 511+t4dt=41.7133..., what is 511+u4du?

139) Estimate 10tdt using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value 10tdt?

Solution: The average is 1/2, which is equal to the integral in this case.

140) Estimate 10tdt by comparison with the area of a single rectangle with height equal to the value of t at the midpoint t=12. How does this midpoint estimate compare with the actual value 10tdt?

141) From the graph of sin(2πx) shown:

CNX_Calc_Figure_05_02_207.jpeg

a. Explain why 10sin(2πt)dt=0.

b. Explain why, in general, a+1asin(2πt)dt=0 for any value of a.

Answer:

a. The graph is antisymmetric with respect to t=12 over [0,1], so the average value is zero.
b. For any value of a, the graph between [a,a+1] is a shift of the graph over [0,1], so the net areas above and below the axis do not change and the average remains zero.

142) If f is 1-periodic (f(t+1)=f(t)), odd, and integrable over [0,1], is it always true that 10f(t)dt=0?

143) If f is 1-periodic and 10f(t)dt=A, is it necessarily true that 1+aaf(t)dt=A for all A?

Solution: Yes, the integral over any interval of length 1 is the same.

J5.2.1) not here yet


5.5: Definite Integral Intro Exercises is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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